HSC 12th Maths June 2026 Board Paper Solution (J-276) | Class 12 Mathematics & Statistics Answer Key

Are you looking for the complete, step-by-step solutions for the HSC 12th Standard Mathematics and Statistics June 2026 Board Question Paper (Subject Code: J-276) held on 24/06/2026? Below is the full, solved answer key covering Section A, Section B, Section C, and Section D prepared strictly according to the Maharashtra State Board marking scheme to help Class 12 students score full marks.

HSC 12th Maths (Arts & Science) June 2026 Board Paper Solution

SECTION - A

Q. 1. Select and write the correct answer of the following multiple choice type of questions:

1. If A = {1, 2, 3, 4, 5} then which of the following is not true?

   Solution: (c) $x \in A, x+6 \ge 9$

   Explanation: If we take $x = 1$ from set A, $1 + 6 = 7$, which is not greater than or equal to 9. Thus, the statement is false.

2. The value of $\sin[\sin^{-1}(-\frac{1}{\sqrt{2}}) - 3 \sin^{-1}(\frac{\sqrt{3}}{2})] =$

   Solution: (b) $\frac{1}{\sqrt{2}}$

   Explanation:
   $\sin^{-1}(-\frac{1}{\sqrt{2}}) = -\frac{\pi}{4}$
   $\sin^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{3}$
   Expression becomes: $\sin(-\frac{\pi}{4} - 3(\frac{\pi}{3})) = \sin(-\frac{\pi}{4} - \pi) = \sin(-\frac{5\pi}{4}) = \frac{1}{\sqrt{2}}$.

3. If $\hat{p}, \hat{q}, \hat{r}$ are mutually perpendicular unit vectors and form a right handed triplet then the value of $\hat{p}\cdot(\hat{q}\times\hat{r}) + \hat{q}\cdot(\hat{p}\times\hat{r}) + \hat{r}\cdot(\hat{p}\times\hat{q}) = $

   Solution: (c) 1

   Explanation: For a right-handed triplet of mutually perpendicular unit vectors:
   $\hat{q} \times \hat{r} = \hat{p}$ (so $\hat{p}\cdot\hat{p} = 1$)
   $\hat{p} \times \hat{r} = -\hat{q}$ (so $\hat{q}\cdot(-\hat{q}) = -1$)
   $\hat{p} \times \hat{q} = \hat{r}$ (so $\hat{r}\cdot\hat{r} = 1$)
   Sum = $1 - 1 + 1 = 1$.

4. The Cartesian equation of the line passing through the points A(3, 4, -1) and B(2, -1, 3) is:

   Solution: (a) $\frac{x-3}{1} = \frac{y-4}{5} = \frac{z+1}{-4}$

   Explanation: Direction ratios (DRs) of AB = (2-3, -1-4, 3-(-1)) = (-1, -5, 4). Equivalent DRs are (1, 5, -4). Using point A(3,4,-1), the equation is $\frac{x-3}{1} = \frac{y-4}{5} = \frac{z+1}{-4}$.

5. If $y = \log_x a$ then $\frac{dy}{dx} =$

   Solution: (a) $\frac{-\log a}{x(\log x)^2}$

   Explanation: Using the change of base rule, $y = \frac{\log a}{\log x} = \log a \cdot (\log x)^{-1}$. Differentiating with respect to $x$: $\frac{dy}{dx} = \log a \cdot (-1)(\log x)^{-2} \cdot \frac{d}{dx}(\log x) = \log a \cdot \frac{-1}{(\log x)^2} \cdot \frac{1}{x} = \frac{-\log a}{x(\log x)^2}$.

6. If $f(x) = \frac{\sin^{-1}x}{\sqrt{1-x^2}}$ and $g(x) = e^{\sin^{-1}x}$ then $\int f(x)\cdot g(x)dx = $

   Solution: (b) $e^{\sin^{-1}x}(\sin^{-1}x - 1) + c$

   Explanation: We need to evaluate $\int \frac{\sin^{-1}x}{\sqrt{1-x^2}} e^{\sin^{-1}x} dx$. Let $t = \sin^{-1}x$, then $dt = \frac{1}{\sqrt{1-x^2}} dx$. The integral becomes $\int t e^t dt$. Using integration by parts: $t \int e^t dt - \int (1 \cdot e^t) dt = t e^t - e^t + c = e^t(t - 1) + c$. Substituting back $t = \sin^{-1}x$, we get $e^{\sin^{-1}x}(\sin^{-1}x - 1) + c$.

7. The area bounded by the line $y=2x$, X-axis and the lines $x=-1$, $x=4$ is _____ sq. units.

   Solution: (b) 17

   Explanation: The area is given by $\int_{-1}^{4} |y| dx = \int_{-1}^{0} |2x| dx + \int_{0}^{4} |2x| dx$. Since $2x$ is negative for $x < 0$, we have $\int_{-1}^{0} (-2x) dx + \int_{0}^{4} (2x) dx$. Evaluating the integrals: $[-x^2]_{-1}^{0} + [x^2]_{0}^{4} = (0 - (-1)) + (16 - 0) = 1 + 16 = 17$ sq. units.

8. In a Binomial Distribution, $E(X)=6$, $Var(X)=4.2$ then the value of n is _____

   Solution: 20

   Explanation: For a Binomial Distribution, Mean $E(X) = np = 6$ and Variance $Var(X) = npq = 4.2$. Dividing variance by mean: $\frac{npq}{np} = \frac{4.2}{6} \Rightarrow q = 0.7$. Since $p = 1 - q$, we get $p = 1 - 0.7 = 0.3$. Substituting $p$ back into the mean equation: $n(0.3) = 6 \Rightarrow n = \frac{6}{0.3} = 20$.

Q. 2. Answer the following questions:

1. Write the dual of $p \wedge \sim q$

   Solution: $p \vee \sim q$

   Explanation: To write the dual of a logical statement, we replace the conjunction symbol ($\wedge$) with the disjunction symbol ($\vee$) and vice versa. Negations ($\sim$) remain unchanged.

2. What is the distance of point (3, 5, 6) from XZ plane?

   Solution: 5 units

   Explanation: The perpendicular distance of any point $(x, y, z)$ from the XZ plane is given by the absolute value of its y-coordinate, $|y|$. Here, the y-coordinate is 5, so the distance is $|5| = 5$ units.

3. The displacement of a particle at time t is given by $S = 3t^2 + 4t + 1$. Find its velocity at time t.

   Solution: $v = 6t + 4$

   Explanation: Velocity ($v$) is the rate of change of displacement ($S$) with respect to time ($t$).
   $v = \frac{dS}{dt} = \frac{d}{dt}(3t^2 + 4t + 1) = 6t + 4$.

4. Write the order of differential equation $(\frac{d^2y}{dx^2})^{\frac{5}{2}} = \sqrt[3]{\frac{dy}{dx}}$

   Solution: Order = 2

   Explanation: The order of a differential equation is the order of the highest derivative present in the equation. Here, the highest derivative is $\frac{d^2y}{dx^2}$, which is a second-order derivative. Therefore, the order is 2.

SECTION - B

Attempt any EIGHT of the following questions:

Q. 3. Construct the switching circuit for the statement $(\sim p \wedge q) \vee (p \wedge \sim r)$.

Solution:

Let,
$p$: Switch $S_1$ is closed.
$q$: Switch $S_2$ is closed.
$r$: Switch $S_3$ is closed.
$\sim p$: Switch $S_1'$ is closed.
$\sim r$: Switch $S_3'$ is closed.

The given logical statement pattern is $(\sim p \wedge q) \vee (p \wedge \sim r)$.

• The term $(\sim p \wedge q)$ indicates that switches $S_1'$ and $S_2$ are connected in series.
• The term $(p \wedge \sim r)$ indicates that switches $S_1$ and $S_3'$ are connected in series.
• The disjunction ($\vee$) between the two brackets indicates that these two series branches are connected in parallel to each other.

Q. 4. Find the matrix of cofactors of matrix $\begin{bmatrix} 5 & 2 \\ -1 & 6 \end{bmatrix}$

Solution:

Let the given matrix be $A = \begin{bmatrix} 5 & 2 \\ -1 & 6 \end{bmatrix}$

Here, the elements of the matrix are:
$a_{11} = 5, a_{12} = 2$
$a_{21} = -1, a_{22} = 6$

The cofactors $C_{ij}$ are calculated using the formula $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$.

• $C_{11} = (-1)^{1+1} (6) = 1 \times 6 = 6$
• $C_{12} = (-1)^{1+2} (-1) = -1 \times -1 = 1$
• $C_{21} = (-1)^{2+1} (2) = -1 \times 2 = -2$
• $C_{22} = (-1)^{2+2} (5) = 1 \times 5 = 5$

Therefore, the matrix of cofactors is:

$\begin{bmatrix} C_{11} & C_{12} \\ C_{21} & C_{22} \end{bmatrix} = \begin{bmatrix} 6 & 1 \\ -2 & 5 \end{bmatrix}$

Q. 5. Find the general solution of equation $\cos 5\theta = \frac{1}{2}$.

Solution:

Given equation is $\cos 5\theta = \frac{1}{2}$

We know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$

Therefore, $\cos 5\theta = \cos(\frac{\pi}{3})$

The general solution for $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$, where $n \in Z$.

Here, $x = 5\theta$ and $\alpha = \frac{\pi}{3}$.

$5\theta = 2n\pi \pm \frac{\pi}{3}$, where $n \in Z$

Dividing both sides by 5, we get:

$\theta = \frac{2n\pi}{5} \pm \frac{\pi}{15}$, where $n \in Z$

This is the required general solution.

Q. 6. Find the value of k, if $2x+y=0$ is one of the lines represented by $3x^2+kxy+2y^2=0$.

Solution:

The given combined equation of the lines is:
$3x^2 + kxy + 2y^2 = 0$

Comparing this with the standard homogeneous equation $ax^2 + 2hxy + by^2 = 0$, we get:
$a = 3$, $2h = k$, and $b = 2$

The corresponding auxiliary equation is $bm^2 + 2hm + a = 0$.
Substituting the values of $a$, $2h$, and $b$, we get:
$2m^2 + km + 3 = 0$     ...(i)

We are given that $2x + y = 0$ is one of the lines represented by the combined equation.
Writing this in the slope-intercept form ($y = mx + c$), we get:
$y = -2x$

Therefore, the slope of this line is $m = -2$.

Since this line is a part of the given combined equation, its slope $m = -2$ must satisfy the auxiliary equation (i).

Substitute $m = -2$ in equation (i):
$2(-2)^2 + k(-2) + 3 = 0$
$2(4) - 2k + 3 = 0$
$8 - 2k + 3 = 0$
$11 - 2k = 0$
$2k = 11$
$k = \frac{11}{2}$

Therefore, the value of $k$ is $\frac{11}{2}$.

Q. 7. If $\overline{a}$, $\overline{b}$, $\overline{c}$ are the position vectors of points A, B, C respectively and $10\overline{a} = 7\overline{b} + 3\overline{c}$ then find the ratio in which the point C divides the line segment AB.

Solution:

Given equation is:
$10\overline{a} = 7\overline{b} + 3\overline{c}$

We need to find the ratio in which point C divides the line segment AB. Let's express $\overline{c}$ in terms of $\overline{a}$ and $\overline{b}$.

$3\overline{c} = 10\overline{a} - 7\overline{b}$

Dividing both sides by 3, we get:

$\overline{c} = \frac{10\overline{a} - 7\overline{b}}{3}$

This can be written in the form of the section formula for external division:

$\overline{c} = \frac{-7\overline{b} + 10\overline{a}}{-7 + 10}$

Comparing this with the section formula $\overline{c} = \frac{m\overline{b} + n\overline{a}}{m + n}$, we get:

$m = -7$ and $n = 10$

The ratio is $m : n = -7 : 10$.

Since the ratio is negative, the division is external. We can also compare it directly with the external division formula $\overline{c} = \frac{m\overline{b} - n\overline{a}}{m - n}$:

$\overline{c} = \frac{7\overline{b} - 10\overline{a}}{7 - 10} = \frac{10\overline{a} - 7\overline{b}}{3}$

Here, $m = 7$ and $n = 10$.

Therefore, the point C divides the line segment AB externally in the ratio 7:10.

Q. 8. Find the angle between the lines $\overline{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(2\hat{i}-\hat{j}+\hat{k})$ and $\overline{r}=(\hat{i}+2\hat{j}+3\hat{k})+\mu(\hat{i}+2\hat{j}+\hat{k})$

Solution:

Let the given lines be:
$L_1: \overline{r} = \overline{a}_1 + \lambda\overline{b}_1$
$L_2: \overline{r} = \overline{a}_2 + \mu\overline{b}_2$

Comparing the given equations with standard forms, the direction vectors of the lines are:
$\overline{b}_1 = 2\hat{i} - \hat{j} + \hat{k}$
$\overline{b}_2 = \hat{i} + 2\hat{j} + \hat{k}$

Let $\theta$ be the acute angle between the two lines. The formula for the angle is:
$\cos \theta = \frac{|\overline{b}_1 \cdot \overline{b}_2|}{|\overline{b}_1| |\overline{b}_2|}$

Now, let's find the dot product $\overline{b}_1 \cdot \overline{b}_2$:
$\overline{b}_1 \cdot \overline{b}_2 = (2\hat{i} - \hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + \hat{k})$
$\overline{b}_1 \cdot \overline{b}_2 = (2)(1) + (-1)(2) + (1)(1)$
$\overline{b}_1 \cdot \overline{b}_2 = 2 - 2 + 1 = 1$

Next, find the magnitudes of the direction vectors:
$|\overline{b}_1| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$
$|\overline{b}_2| = \sqrt{(1)^2 + (2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

Substitute these values into the formula:
$\cos \theta = \frac{|1|}{\sqrt{6} \times \sqrt{6}}$
$\cos \theta = \frac{1}{6}$

Therefore, the angle between the lines is:
$\theta = \cos^{-1}(\frac{1}{6})$

Q. 9. Test whether the function $f(x) = x - \frac{1}{x}$, $x \in R, x \ne 0$ is increasing or decreasing.

Solution:

The given function is $f(x) = x - \frac{1}{x}$

Differentiating with respect to $x$, we get:

$f'(x) = \frac{d}{dx}(x) - \frac{d}{dx}(\frac{1}{x})$

$f'(x) = 1 - (-\frac{1}{x^2})$

$f'(x) = 1 + \frac{1}{x^2}$

For any real number $x \ne 0$, the square of $x$ is always positive, i.e., $x^2 > 0$.

Therefore, $\frac{1}{x^2} > 0$

Adding 1 to both sides:

$1 + \frac{1}{x^2} > 1 > 0$

This means $f'(x) > 0$ for all $x \in R, x \ne 0$.

Hence, the given function $f(x)$ is strictly increasing for all $x \in R, x \ne 0$.

Q. 10. Evaluate: $\int_{-1}^{1}|x|dx$

Solution:

Let $I = \int_{-1}^{1}|x|dx$

We know the definition of the absolute value function:

$|x| = -x$ for $x < 0$

$|x| = x$ for $x \ge 0$

We split the definite integral at $x = 0$:

$I = \int_{-1}^{0} (-x) dx + \int_{0}^{1} (x) dx$

Integrating both terms:

$I = \left[ -\frac{x^2}{2} \right]_{-1}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{1}$

Applying the limits:

$I = \left( 0 - \left(-\frac{(-1)^2}{2}\right) \right) + \left( \frac{1^2}{2} - 0 \right)$

$I = \left( 0 - \left(-\frac{1}{2}\right) \right) + \left( \frac{1}{2} \right)$

$I = \frac{1}{2} + \frac{1}{2}$

$I = 1$

(Alternatively, since $f(x) = |x|$ is an even function, $\int_{-1}^{1}|x|dx = 2\int_{0}^{1}xdx = 2 \left[ \frac{x^2}{2} \right]_{0}^{1} = 2(\frac{1}{2}) = 1$)

Question 11

Evaluate: ∫ x log x dx

Solution:

Let I = ∫ x log x dx

We solve this using the method of integration by parts. The formula is:

∫ u · v dx = u ∫ v dx - ∫ [ (du/dx) · ∫ v dx ] dx

According to the LIATE rule (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential), we choose our functions as follows:

• u = log x (Logarithmic function)
• v = x (Algebraic function)

Applying the integration by parts formula:

I = (log x) ∫ x dx - ∫ [ (d/dx (log x)) · ∫ x dx ] dx

Now, compute the individual derivatives and integrals:

• ∫ x dx = x² / 2
• d/dx (log x) = 1 / x

Substitute these back into the equation:

I = (log x) · (x² / 2) - ∫ [ (1 / x) · (x² / 2) ] dx

I = (x² / 2) log x - (1 / 2) ∫ x dx

I = (x² / 2) log x - (1 / 2) · (x² / 2) + C

I = (x² / 2) log x - (x² / 4) + C

Where 'C' is the constant of integration.

Question 12

Find the area bounded by curve x² + y = 0, X-axis and lines x = 1 and x = 4.

Solution:

The equation of the given curve is:

x² + y = 0 ⇒ y = -x²

Required area (A) bounded by the curve, the X-axis, and the lines x = 1 and x = 4 is given by:

A = ∫₁⁴ |y| dx

A = ∫₁⁴ |-x²| dx

A = ∫₁⁴ x² dx

Integrating with respect to x:

A = [ x³ / 3 ]₁⁴

A = (4³ / 3) - (1³ / 3)

A = (64 / 3) - (1 / 3)

A = 63 / 3

A = 21 sq. units

Question 13

The probability distribution of r.v. X is as follows:

X = x	0	1	2	3	4
P(X = x)	0.1	k	2k	2k	k

Find : (i) k (ii) F(2)

Solution:

Since P(X = x) is the probability distribution of a random variable X, the sum of all probabilities is equal to 1:

∑ P(X = x) = 1

(i) To find k:

P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 1

0.1 + k + 2k + 2k + k = 1

0.1 + 6k = 1

6k = 1 - 0.1

6k = 0.9

k = 0.9 / 6

k = 0.15 (or 3/20)

(ii) To find F(2):

By definition of the cumulative distribution function (c.d.f.):

F(2) = P(X ≤ 2)

F(2) = P(X = 0) + P(X = 1) + P(X = 2)

F(2) = 0.1 + k + 2k

F(2) = 0.1 + 3k

Substituting k = 0.15:

F(2) = 0.1 + 3(0.15)

F(2) = 0.1 + 0.45

F(2) = 0.55 (or 11/20)

Question 14

Evaluate: ∫ [ e^x(1 + x) / cos²(xe^x) ] dx

Solution:

Let I = ∫ [ e^x(1 + x) / cos²(xe^x) ] dx

Use the method of substitution by putting:

xe^x = t

Differentiating both sides with respect to x using the product rule:

[ x · (d/dx (e^x)) + e^x · (d/dx (x)) ] dx = dt

(x e^x + e^x · 1) dx = dt

e^x(1 + x) dx = dt

Substitute these values into the integral I:

I = ∫ (1 / cos² t) dt

I = ∫ sec² t dt

I = tan t + C

Resubstituting t = xe^x:

I = tan(xe^x) + C

Where 'C' is the constant of integration.

Section C

Question 15

Problem: Determine using a truth table whether the logical expression [(p ∨ q) ∧ ~p] ∧ ~q is a tautology, a contradiction, or a contingency.

Solution:

p	q	~p	~q	p ∨ q	(p ∨ q) ∧ ~p	[(p ∨ q) ∧ ~p] ∧ ~q
T	T	F	F	T	F	F
T	F	F	T	T	F	F
F	T	T	F	T	T	F
F	F	T	T	F	F	F

All the truth values in the last column are False (F).

Conclusion: The given statement pattern is a contradiction.

Question 16

Problem: Show that sin^-1(3/5) + cos^-1(12/13) = sin^-1(56/65).

Solution:

Let sin^-1(3/5) = x and cos^-1(12/13) = y, where 0 < x, y < π/2.

Therefore, sin x = 3/5 and cos y = 12/13.

Using the trigonometric identity sin²θ + cos²θ = 1:

• cos x = √(1 - sin²x) = √(1 - (3/5)²) = √(1 - 9/25) = √(16/25) = 4/5
• sin y = √(1 - cos²y) = √(1 - (12/13)²) = √(1 - 144/169) = √(25/169) = 5/13

We know the compound angle formula:

sin(x + y) = sin x cos y + cos x sin y

sin(x + y) = (3/5)(12/13) + (4/5)(5/13)

sin(x + y) = (36/65) + (20/65)

sin(x + y) = 56/65

x + y = sin^-1(56/65)

Resubstituting the values of x and y:

sin^-1(3/5) + cos^-1(12/13) = sin^-1(56/65)

Hence proved.

Question 17

Problem: Given a triangle ABC with sides a = 10, b = 8, and c = 6, compute the values of (i) sin(C/2) and (ii) cos(B/2).

Solution:

Given: a = 10, b = 8, c = 6

First, find the semi-perimeter (s) of ΔABC:

2s = a + b + c = 10 + 8 + 6 = 24

s = 12

Now compute:

• s - a = 12 - 10 = 2
• s - b = 12 - 8 = 4
• s - c = 12 - 6 = 6

(i) To find sin(C/2):

By the half-angle formula:

sin(C/2) = √[ (s - a)(s - b) / (ab) ]

sin(C/2) = √[ (2 × 4) / (10 × 8) ]

sin(C/2) = √(8 / 80) = √(1 / 10)

sin(C/2) = 1 / √10

(ii) To find cos(B/2):

By the half-angle formula:

cos(B/2) = √[ s(s - b) / (ac) ]

cos(B/2) = √[ (12 × 4) / (10 × 6) ]

cos(B/2) = √(48 / 60) = √(4 / 5)

cos(B/2) = 2 / √5

Question 18

Problem: Determine with justification whether the four points A(1, 2, 1), B(2, -3, 4), C(3, 4, -5), and D(2, 3, -2) lie on the same plane.

Solution:

Let the position vectors of the given points be:

• a = î + 2ĵ + k̂
• b = 2î - 3ĵ + 4k̂
• c = 3î + 4ĵ - 5k̂
• d = 2î + 3ĵ - 2k̂

Find the vectors AB, AC, and AD:

• AB = b - a = (2 - 1)î + (-3 - 2)ĵ + (4 - 1)k̂ = î - 5ĵ + 3k̂
• AC = c - a = (3 - 1)î + (4 - 2)ĵ + (-5 - 1)k̂ = 2î + 2ĵ - 6k̂
• AD = d - a = (2 - 1)î + (3 - 2)ĵ + (-2 - 1)k̂ = î + ĵ - 3k̂

Now, evaluate the scalar triple product [AB AC AD]:

| 1	-5	3 |
| 2	2	-6 |
| 1	1	-3 |

[AB AC AD] = 1( (2)(-3) - (-6)(1) ) - (-5)( (2)(-3) - (-6)(1) ) + 3( (2)(1) - (2)(1) )

[AB AC AD] = 1(-6 + 6) + 5(-6 + 6) + 3(2 - 2)

[AB AC AD] = 1(0) + 5(0) + 3(0) = 0

Since the scalar triple product [AB AC AD] = 0, the vectors AB, AC, and AD are coplanar.

Conclusion: Yes, the four points A, B, C, and D are coplanar.

Question 19

Problem: Calculate the shortest distance between the parallel lines x/2 = y/(-1) = z/2 and (x + 1)/2 = (y - 1)/(-1) = (z + 1)/2.

Solution:

The equations of the given parallel lines are:

• Line 1: x/2 = y/(-1) = z/2
• Line 2: (x + 1)/2 = (y - 1)/(-1) = (z + 1)/2

From the equations of the lines:

• Position vector of a point on the first line: a₁ = 0î + 0ĵ + 0k̂
• Position vector of a point on the second line: a₂ = -î + ĵ - k̂
• Direction vector parallel to both lines: b = 2î - ĵ + 2k̂

Now, compute a₂ - a₁:

a₂ - a₁ = -î + ĵ - k̂

Next, compute the cross product (a₂ - a₁) × b:

| î	ĵ	k̂ |
| -1	1	-1 |
| 2	-1	2 |

(a₂ - a₁) × b = î(2 - 1) - ĵ(-2 + 2) + k̂(1 - 2)

(a₂ - a₁) × b = î - 0ĵ - k̂ = î - k̂

Find the magnitudes:

• |(a₂ - a₁) × b| = √(1² + 0² + (-1)²) = √2
• |b| = √(2² + (-1)² + 2²) = √(4 + 1 + 4) = √9 = 3

The distance (d) between the two parallel lines is given by:

d = |(a₂ - a₁) × b| / |b|

d = (√2 / 3) units

Question 20

Problem: Determine the vector equation of the plane passing through the points A(2, 1, 1), B(0, 2, 3), and C(4, 5, 6).

Solution:

Let the position vectors of points A, B, and C be a, b, and c respectively:

• a = 2î + ĵ + k̂
• b = 0î + 2ĵ + 3k̂
• c = 4î + 5ĵ + 6k̂

Then:

• AB = b - a = (0 - 2)î + (2 - 1)ĵ + (3 - 1)k̂ = -2î + ĵ + 2k̂
• AC = c - a = (4 - 2)î + (5 - 1)ĵ + (6 - 1)k̂ = 2î + 4ĵ + 5k̂

A normal vector to the plane is given by n = AB × AC:

| î	ĵ	k̂ |
| -2	1	2 |
| 2	4	5 |

AB × AC = î(5 - 8) - ĵ(-10 - 4) + k̂(-8 - 2)

AB × AC = -3î + 14ĵ - 10k̂

Now calculate a · (AB × AC):

a · (AB × AC) = (2î + ĵ + k̂) · (-3î + 14ĵ - 10k̂)

a · (AB × AC) = 2(-3) + 1(14) + 1(-10) = -6 + 14 - 10 = -2

The vector equation of the plane passing through three points is:

r · (AB × AC) = a · (AB × AC)

r · (-3î + 14ĵ - 10k̂) = -2

Or equivalently:

r · (3î - 14ĵ + 10k̂) = 2

Question 21

Problem: Given y = log[ (x + √(x² + a²)) / (√(x² + a²) - x) ], find dy/dx.

Solution:

Given:

y = log[ (x + √(x² + a²)) / (√(x² + a²) - x) ]

Using the logarithmic property log(m / n) = log m - log n:

y = log(x + √(x² + a²)) - log(√(x² + a²) - x)

Differentiating both sides with respect to x:

dy/dx = [ 1 / (x + √(x² + a²)) ] · d/dx (x + √(x² + a²)) - [ 1 / (√(x² + a²) - x) ] · d/dx (√(x² + a²) - x)

dy/dx = [ 1 / (x + √(x² + a²)) ] · [ 1 + (2x / (2√(x² + a²))) ] - [ 1 / (√(x² + a²) - x) ] · [ (2x / (2√(x² + a²))) - 1 ]

dy/dx = [ 1 / (x + √(x² + a²)) ] · [ (√(x² + a²) + x) / √(x² + a²) ] - [ 1 / (√(x² + a²) - x) ] · [ (x - √(x² + a²)) / √(x² + a²) ]

dy/dx = [ 1 / √(x² + a²) ] - [ -(√(x² + a²) - x) / ( (√(x² + a²) - x) √(x² + a²) ) ]

dy/dx = [ 1 / √(x² + a²) ] + [ 1 / √(x² + a²) ]

dy/dx = 2 / √(x² + a²)

Question 22

Problem: A wire of total length 36 cm is bent into the shape of a rectangle. Find the dimensions of the rectangle so that its enclosed area is maximized.

Solution:

Let the length of the rectangle be x cm and the breadth of the rectangle be y cm.

Given that the perimeter of the rectangle is equal to the length of the wire (36 cm):

2(x + y) = 36

x + y = 18

y = 18 - x — (Equation 1)

Let A be the area of the rectangle:

A = x · y

A = x(18 - x)

A = 18x - x²

Differentiating A with respect to x:

dA/dx = 18 - 2x

Differentiating again with respect to x:

d²A/dx² = -2

For maximum or minimum area, put dA/dx = 0:

18 - 2x = 0

2x = 18 ⇒ x = 9

At x = 9, d²A/dx² = -2 < 0.

By the second derivative test, the area A is maximum when x = 9 cm.

Substituting x = 9 in Equation 1:

y = 18 - 9 ⇒ y = 9

Conclusion: For the area to be maximum, the dimensions of the rectangle are length = 9 cm and breadth = 9 cm.

Question 23

Problem: Verify Rolle's theorem for the polynomial function f(x) = x² - 5x + 9 on the interval [1, 4].

Solution:

Given function: f(x) = x² - 5x + 9 on x ∈ [1, 4].

Since f(x) is a polynomial function:

1. f(x) is continuous on the closed interval [1, 4].
2. f(x) is differentiable on the open interval (1, 4).

Now check the values of f(x) at the endpoints of the interval:

• f(1) = (1)² - 5(1) + 9 = 1 - 5 + 9 = 5
• f(4) = (4)² - 5(4) + 9 = 16 - 20 + 9 = 5

Here, f(1) = f(4) = 5.

Thus, all three conditions of Rolle's theorem are satisfied. Therefore, there exists at least one real number c ∈ (1, 4) such that f'(c) = 0.

Differentiating f(x) with respect to x:

f'(x) = 2x - 5

Put f'(c) = 0:

2c - 5 = 0

2c = 5 ⇒ c = 5/2 = 2.5

Since c = 2.5 ∈ (1, 4), Rolle's theorem is verified.

Question 24

Problem: Find the general solution of the differential equation x(dy/dx) - √(x² + y²) = y.

Solution:

Given differential equation:

x(dy/dx) - √(x² + y²) = y

x(dy/dx) = y + √(x² + y²)

dy/dx = [ y + √(x² + y²) ] / x — (Equation 1)

This is a homogeneous differential equation. Put y = vx.

Differentiating with respect to x:

dy/dx = v + x(dv/dx)

Substitute y = vx and dy/dx into Equation 1:

v + x(dv/dx) = [ vx + √(x² + v²x²) ] / x

v + x(dv/dx) = [ vx + x√(1 + v²) ] / x

v + x(dv/dx) = v + √(1 + v²)

x(dv/dx) = √(1 + v²)

Separating the variables:

[ 1 / √(1 + v²) ] dv = (1 / x) dx

Integrating both sides:

∫ [ 1 / √(1 + v²) ] dv = ∫ (1 / x) dx

log|v + √(1 + v²)| = log|x| + log|c|

log|v + √(1 + v²)| = log|cx|

v + √(1 + v²) = cx

Resubstituting v = y / x:

(y / x) + √(1 + y²/x²) = cx

[ y + √(x² + y²) ] / x = cx

y + √(x² + y²) = cx²

This is the required general solution.

Question 25

Problem: For a continuous random variable X with probability density function f(x) = x/8 for 0 < x < 4 (and 0 otherwise), find (i) the cumulative distribution function F(x), and (ii) the values of F(x) at x = 0.5 and x = 5.

Solution:

Given p.d.f. of continuous random variable X:

• f(x) = x / 8, for 0 < x < 4
• f(x) = 0, otherwise

(i) Expression for c.d.f. F(x):

By definition, F(x) = ∫_-∞^x f(t) dt.

• For x ≤ 0: F(x) = 0
• For 0 < x < 4:

  F(x) = ∫₀^x (t / 8) dt = (1 / 8) [ t² / 2 ]₀^x = x² / 16

• For x ≥ 4: F(x) = 1

(ii) Values of F(x) at x = 0.5 and x = 5:

• At x = 0.5: Since 0 < 0.5 < 4, we use F(x) = x² / 16:

  F(0.5) = (0.5)² / 16 = 0.25 / 16 = (1/4) / 16 = 1 / 64 (or 0.015625)

• At x = 5: Since x = 5 ≥ 4, the cumulative probability reaches its maximum value:

  F(5) = 1

Question 26

Problem: Given the probability mass function of a discrete random variable X as P(X = x) = ⁴C_x (5/9)^x (4/9)^4-x for x = 0, 1, 2, 3, 4 (and 0 otherwise), calculate the expected value E(X) and variance Var(X).

Solution:

Given p.m.f.:

P(X = x) = ⁴C_x (5/9)^x (4/9)^4-x, for x = 0, 1, 2, 3, 4

Comparing this with the standard Binomial distribution formula P(X = x) = ⁿC_x p^x q^n-x, we identify that X follows a Binomial distribution B(n, p) with:

• Number of trials, n = 4
• Probability of success, p = 5/9
• Probability of failure, q = 4/9

1. To find E(X):

For a binomial distribution, the mean or expected value is given by:

E(X) = n · p

E(X) = 4 × (5/9)

E(X) = 20 / 9

2. To find Var(X):

For a binomial distribution, the variance is given by:

Var(X) = n · p · q

Var(X) = 4 × (5/9) × (4/9)

Var(X) = 80 / 81

Section D

Question 27

Problem: Find the inverse of the matrix:

[ -4	-3	-3 ]
[ 1	0	1 ]
[ 4	4	3 ]

Solution:

Let the given matrix be A:

[ -4	-3	-3 ]
[ 1	0	1 ]
[ 4	4	3 ]

Step 1: Find the determinant |A|

|A| = -4(0 × 3 - 1 × 4) - (-3)(1 × 3 - 1 × 4) + (-3)(1 × 4 - 0 × 4)

|A| = -4(0 - 4) + 3(3 - 4) - 3(4 - 0)

|A| = -4(-4) + 3(-1) - 3(4)

|A| = 16 - 3 - 12 = 1 ≠ 0

Since |A| ≠ 0, the matrix A is non-singular and A^-1 exists.

Step 2: Find the cofactors of the elements of matrix A

• A₁₁ = (-1)¹⁺¹ (0 - 4) = -4
• A₁₂ = (-1)¹⁺² (3 - 4) = -(-1) = 1
• A₁₃ = (-1)¹⁺³ (4 - 0) = 4
• A₂₁ = (-1)²⁺¹ (-9 - (-12)) = -(3) = -3
• A₂₂ = (-1)²⁺² (-12 - (-12)) = 0
• A₂₃ = (-1)²⁺³ (-16 - (-12)) = -(-4) = 4
• A₃₁ = (-1)³⁺¹ (-3 - 0) = -3
• A₃₂ = (-1)³⁺² (-4 - (-3)) = -(-1) = 1
• A₃₃ = (-1)³⁺³ (0 - (-3)) = 3

The matrix of cofactors [A_ij] is:

[ -4	1	4 ]
[ -3	0	4 ]
[ -3	1	3 ]

Step 3: Find the adjoint of matrix A

adj(A) is the transpose of the cofactor matrix:

[ -4	-3	-3 ]
[ 1	0	1 ]
[ 4	4	3 ]

Step 4: Find the inverse A^-1

Using the formula A^-1 = (1 / |A|) · adj(A):

A^-1 = (1 / 1) · adj(A)

Therefore, A^-1 is:

[ -4	-3	-3 ]
[ 1	0	1 ]
[ 4	4	3 ]

Question 28

Problem: Prove that the homogeneous equation of degree two in x and y, ax² + 2hxy + by² = 0, represents a pair of lines passing through the origin if h² - ab ≥ 0.

Solution:

Consider the homogeneous equation of degree two in x and y:

ax² + 2hxy + by² = 0 — (Equation 1)

In this second-degree equation, the coefficients a, h, and b cannot all be zero simultaneously. We consider two cases based on the value of b:

Case 1: When b = 0

If b = 0, then a and h cannot both be zero. Substituting b = 0 into Equation 1 gives:

ax² + 2hxy = 0

x(ax + 2hy) = 0

This gives two linear equations:

x = 0 and ax + 2hy = 0

Since both are first-degree equations in x and y with no constant term, they represent two straight lines passing through the origin.

Case 2: When b ≠ 0

Multiply both sides of Equation 1 by b:

abx² + 2hbxy + b²y² = 0

b²y² + 2hbxy = -abx²

To complete the square on the left-hand side, add h²x² to both sides:

b²y² + 2hbxy + h²x² = h²x² - abx²

(by + hx)² = (h² - ab)x²

Given that h² - ab ≥ 0, we can write (h² - ab) = [√(h² - ab)]²:

(hx + by)² - [x√(h² - ab)]² = 0

Factoring using a² - b² = (a + b)(a - b):

[ (hx + by) + x√(h² - ab) ] [ (hx + by) - x√(h² - ab) ] = 0

[ (h + √(h² - ab))x + by ] [ (h - √(h² - ab))x + by ] = 0

This equation is the combined equation of the two lines:

• (h + √(h² - ab))x + by = 0
• (h - √(h² - ab))x + by = 0

Since both equations are linear in x and y and lack constant terms, they represent a pair of straight lines passing through the origin.

Hence proved.

Question 29

Problem: Prove that, if a, b, c are three non-coplanar vectors, then any vector r in space can be uniquely expressed as a linear combination of a, b, c.

Solution:

Part 1: Existence of the Linear Combination

Let a, b, c be three non-coplanar vectors in space, and let r be any arbitrary vector in space.

Take a fixed point O in space as the origin. Let the vectors a, b, c, and r be positioned along the directed line segments from O such that OP = r, and the lines along a, b, and c pass through O.

Through point P, draw planes parallel to the planes determined by the pairs of vectors (b, c), (c, a), and (a, b). Let these planes intersect the lines containing a, b, and c at points A, B, and C respectively, forming a parallelepiped with OP as its diagonal and OA, OB, OC as its coterminous edges.

• Since OA is collinear with a, there exists a scalar x ∈ R such that OA = xa.
• Since OB is collinear with b, there exists a scalar y ∈ R such that OB = yb.
• Since OC is collinear with c, there exists a scalar z ∈ R such that OC = zc.

By the parallelepiped law of vector addition:

OP = OA + OB + OC

r = xa + yb + zc

Thus, the vector r can be expressed as a linear combination of a, b, and c.

Part 2: Uniqueness of the Representation

Suppose the vector r can also be expressed as another linear combination of a, b, and c using scalars x', y', z':

r = x'a + y'b + z'c

Equating the two expressions for r:

xa + yb + zc = x'a + y'b + z'c

(x - x')a + (y - y')b + (z - z')c = 0

Since a, b, and c are three non-coplanar vectors, they are linearly independent. If any scalar coefficient were non-zero (for example, if x - x' ≠ 0), we could express a as a linear combination of b and c, which would contradict the fact that a, b, and c are non-coplanar.

Therefore, all coefficients must equal zero:

• x - x' = 0 ⇒ x = x'
• y - y' = 0 ⇒ y = y'
• z - z' = 0 ⇒ z = z'

Hence, the linear combination is unique.

Hence proved.

Question 30

Problem: Solve the linear programming problem (L.P.P.) by graphical method:

Minimize: z = 8x + 10y

Subject to:

• 2x + y ≥ 7
• 2x + 3y ≥ 15
• x ≥ 0
• y ≥ 2

Solution:

Step 1: Convert the given inequations into equations to plot the boundary lines

Inequation	Equation of Line	Points on the Line	Origin Test (0, 0)	Feasible Half-Plane
2x + y ≥ 7	2x + y = 7 (L1)	(0, 7), (3.5, 0)	0 ≥ 7 (False)	Non-origin side of L1
2x + 3y ≥ 15	2x + 3y = 15 (L2)	(0, 5), (7.5, 0)	0 ≥ 15 (False)	Non-origin side of L2
y ≥ 2	y = 2 (L3)	(0, 2), line parallel to X-axis	0 ≥ 2 (False)	Non-origin side (above y = 2)
x ≥ 0	x = 0 (Y-axis)	Along Y-axis	—	Right side of Y-axis

Step 2: Determine the corner points of the feasible region

Plotting the lines and shading the common region in the first quadrant gives an unbounded feasible region with vertices A, B, and C:

1. Point A: Intersection of x = 0 and 2x + y = 7:

   Substituting x = 0 gives y = 7. Thus, A = (0, 7).

2. Point B: Intersection of 2x + y = 7 and 2x + 3y = 15:

   Subtracting (2x + y = 7) from (2x + 3y = 15) gives 2y = 8 ⇒ y = 4.

   Substituting y = 4 into 2x + y = 7 gives 2x + 4 = 7 ⇒ 2x = 3 ⇒ x = 1.5 (or 3/2).

   Thus, B = (1.5, 4).

3. Point C: Intersection of 2x + 3y = 15 and y = 2:

   Substituting y = 2 into 2x + 3y = 15 gives 2x + 6 = 15 ⇒ 2x = 9 ⇒ x = 4.5 (or 9/2).

   Thus, C = (4.5, 2).

Step 3: Evaluate the objective function z = 8x + 10y at the corner points

Corner Point	Coordinates (x, y)	Value of z = 8x + 10y
A	(0, 7)	z = 8(0) + 10(7) = 0 + 70 = 70
B	(1.5, 4)	z = 8(1.5) + 10(4) = 12 + 40 = 52 (Minimum)
C	(4.5, 2)	z = 8(4.5) + 10(2) = 36 + 20 = 56

Conclusion: The minimum value of z is 52, which occurs at the point x = 1.5 (or 3/2) and y = 4.

Question 31

Problem: If x = f(t) and y = g(t) are differentiable functions of t, so that y is a differentiable function of x and dx/dt ≠ 0, then prove that dy/dx = (dy/dt) / (dx/dt). Hence find dy/dx, if x = sin t and y = cos t.

Solution:

Part 1: Proof of Parametric Differentiation Theorem

Let x = f(t) and y = g(t) be differentiable functions of a parameter t.

Let δt be a small increment in t, and let δx and δy be the corresponding small increments in x and y respectively, such that δx ≠ 0. As δt → 0, we have δx → 0 and δy → 0.

Since x and y are differentiable functions of t:

lim_{δt → 0} (δx / δt) = dx/dt and lim_{δt → 0} (δy / δt) = dy/dt

Consider the increment ratio:

δy / δx = (δy / δt) / (δx / δt), where (δx / δt) ≠ 0

Taking the limit as δt → 0 on both sides:

lim_{δt → 0} (δy / δx) = [ lim_{δt → 0} (δy / δt) ] / [ lim_{δt → 0} (δx / δt) ]

Since δx → 0 as δt → 0, and y is a differentiable function of x, the left-hand side limit exists and equals dy/dx:

lim_{δx → 0} (δy / δx) = (dy/dt) / (dx/dt)

dy/dx = (dy/dt) / (dx/dt), where dx/dt ≠ 0

Hence proved.

Part 2: To find dy/dx when x = sin t and y = cos t

Given: x = sin t and y = cos t

Differentiating both with respect to t:

• dx/dt = d/dt (sin t) = cos t
• dy/dt = d/dt (cos t) = -sin t

Using the parametric differentiation formula:

dy/dx = (dy/dt) / (dx/dt)

dy/dx = -sin t / cos t

dy/dx = -tan t

Question 32

Problem: Evaluate: ∫ [ 2 / (3 + 2sin²x + 5cos²x) ] dx

Solution:

Let I = ∫ [ 2 / (3 + 2sin²x + 5cos²x) ] dx

Divide both the numerator and the denominator by cos²x:

I = ∫ [ (2 / cos²x) / ( (3 / cos²x) + 2(sin²x / cos²x) + 5(cos²x / cos²x) ) ] dx

I = ∫ [ 2sec²x / (3sec²x + 2tan²x + 5) ] dx

Use the trigonometric identity sec²x = 1 + tan²x in the denominator:

I = ∫ [ 2sec²x / (3(1 + tan²x) + 2tan²x + 5) ] dx

I = ∫ [ 2sec²x / (3 + 3tan²x + 2tan²x + 5) ] dx

I = ∫ [ 2sec²x / (5tan²x + 8) ] dx

Substitute tan x = t.

Differentiating with respect to x gives: sec²x dx = dt.

Substitute into the integral I:

I = ∫ [ 2 / (5t² + 8) ] dt

I = (2 / 5) ∫ [ 1 / (t² + 8/5) ] dt

I = (2 / 5) ∫ [ 1 / (t² + (√(8/5))²) ] dt

Using the standard integration formula ∫ [ 1 / (x² + a²) ] dx = (1 / a) tan^-1(x / a) + C:

I = (2 / 5) · [ 1 / √(8/5) ] tan^-1[ t / √(8/5) ] + C

I = (2 / 5) · ( √5 / 2√2 ) tan^-1( √5 t / 2√2 ) + C

I = ( 1 / √10 ) tan^-1( √5 t / 2√2 ) + C

Resubstituting t = tan x:

I = ( 1 / √10 ) tan^-1[ ( √5 tan x ) / ( 2√2 ) ] + C

Where 'C' is the constant of integration.

Question 33

Problem: Prove that:

∫_-a^a f(x) dx = 2 ∫₀^a f(x) dx, if f(x) is an even function

∫_-a^a f(x) dx = 0, if f(x) is an odd function

Solution:

Since -a < 0 < a, by the additive property of definite integrals, we can split the integral at 0:

∫_-a^a f(x) dx = ∫_-a⁰ f(x) dx + ∫₀^a f(x) dx — (Equation 1)

Consider the first integral on the right-hand side:

I₁ = ∫_-a⁰ f(x) dx

Put x = -t. Differentiating with respect to t gives dx = -dt.

Changing the limits of integration:

• When x = -a, t = a
• When x = 0, t = 0

Substituting into I₁:

I₁ = ∫_a⁰ f(-t) (-dt) = -∫_a⁰ f(-t) dt

Using the property -∫_a^b g(t) dt = ∫_b^a g(t) dt:

I₁ = ∫₀^a f(-t) dt

Since a definite integral is independent of the variable of integration (∫₀^a f(-t) dt = ∫₀^a f(-x) dx):

I₁ = ∫₀^a f(-x) dx

Substituting I₁ back into Equation 1:

∫_-a^a f(x) dx = ∫₀^a f(-x) dx + ∫₀^a f(x) dx

∫_-a^a f(x) dx = ∫₀^a [ f(-x) + f(x) ] dx — (Equation 2)

Case 1: If f(x) is an even function

By definition of an even function, f(-x) = f(x). Substituting this into Equation 2:

∫_-a^a f(x) dx = ∫₀^a [ f(x) + f(x) ] dx

∫_-a^a f(x) dx = 2 ∫₀^a f(x) dx

Case 2: If f(x) is an odd function

By definition of an odd function, f(-x) = -f(x). Substituting this into Equation 2:

∫_-a^a f(x) dx = ∫₀^a [ -f(x) + f(x) ] dx

∫_-a^a f(x) dx = 0

Hence proved.

Question 34

Problem: A body cools according to Newton's law from 100°C to 60°C in 20 minutes. The temperature of the surrounding being 20°C. How long will it take to cool down to 40°C?

Solution:

Let θ be the temperature of the body at time t minutes. The temperature of the surrounding medium is θ₀ = 20°C.

According to Newton's Law of Cooling, the rate of change of temperature is directly proportional to the difference between the temperature of the body and the surrounding medium:

dθ/dt ∝ -(θ - θ₀)

dθ/dt = -k(θ - 20), where k > 0 is the constant of proportionality.

Separating the variables:

[ 1 / (θ - 20) ] dθ = -k dt

Integrating both sides:

∫ [ 1 / (θ - 20) ] dθ = -k ∫ dt

log(θ - 20) = -kt + c — (Equation 1)

Step 1: Find the constant c using the initial condition

Initially, at t = 0, θ = 100°C. Substituting into Equation 1:

log(100 - 20) = -k(0) + c

c = log 80

Substituting c = log 80 back into Equation 1:

log(θ - 20) = -kt + log 80

log(θ - 20) - log 80 = -kt

log[ (θ - 20) / 80 ] = -kt — (Equation 2)

Step 2: Find the value of -k using the second condition

Given that at t = 20 minutes, θ = 60°C. Substituting into Equation 2:

log[ (60 - 20) / 80 ] = -20k

log(40 / 80) = -20k

log(1 / 2) = -20k

-k = (1 / 20) log(1 / 2)

Substituting -k into Equation 2 gives:

log[ (θ - 20) / 80 ] = (t / 20) log(1 / 2) — (Equation 3)

Step 3: Find time t when θ = 40°C

Substitute θ = 40 into Equation 3:

log[ (40 - 20) / 80 ] = (t / 20) log(1 / 2)

log(20 / 80) = (t / 20) log(1 / 2)

log(1 / 4) = (t / 20) log(1 / 2)

log[ (1 / 2)² ] = (t / 20) log(1 / 2)

2 log(1 / 2) = (t / 20) log(1 / 2)

Dividing both sides by log(1 / 2):

2 = t / 20

t = 40 minutes

Conclusion: It will take 40 minutes in total for the body to cool down to 40°C.

HSC 12th Maths June 2026 Board Question Paper (J-276) – Original Images

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HSC Accounts March Board Paper With Solution.

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ACCOUNTS BOARD QUESTION PAPER: WITH COMPLETE SOLUTION

SOLUTION TO BK & ACCOUNTANCY MARCH BOARD BK SOLUTION. IMPORTANT FOR BOARD EXAM.

Max. Marks: 80 | Time: 3 Hrs.

Q.1. Attempt all of the following sub-questions:

(A) Select the correct options and rewrite the statements:

1. To find out the net profit or net loss of the business _______ Account is prepared.

(a) Trading (b) Capital (c) Current (d) Profit and Loss

Show Answer

Answer: (d) Profit and Loss

2. From financial statement analysis the creditors are specially interested to know _______.

(a) Liquidity (b) Profits (c) Sale (d) Share Capital

Show Answer

Answer: (a) Liquidity

3. Death is a compulsory _______.

(a) dissolution (b) admission (c) retirement (d) winding up

Show Answer

Answer: (c) retirement

4. The due date of the bill drawn for 2 months on 23rd November, 2019 will be _______.

(a) 23rd Jan, 2020 (b) 25th Jan, 2019 (c) 26th Jan, 2019 (d) 25th Jan, 2020

Show Answer

Answer: (d) 25th Jan, 2020

5. Decrease in the value of assets should be _______ to Profit and Loss Adjustment Account.

(a) debited (b) credited (c) added (d) none of the above

Show Answer

Answer: (a) debited

---

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(B) Write a word / term / phrase as a substitute for each of the following statements: (5)

1. Debit balance of Trading Account.

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Answer: Gross Loss.

2. Expenses incurred on dissolution of the firm.

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Answer: Dissolution/realization Expenses

3. Old Ratio less New Ratio.

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Answer: Sacrifice Ratio

4. Officer appointed by Govt. for noting the dishonor of the bill.

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Answer: Notary Public.

5. Donation received for a specific purpose.

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Answer: Specific donation / Capital receipt.

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(C) Answer the following questions in only ‘one’ sentence each: (5)

1. What is Legacy?

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Solution: Any asset, property or amount of cash which ‘Not for Profit’ concern receives as per the provisions made in the will of the donor after his death is called Legacy.

2. What is CAS?

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Solution: CAS means Computerized Accounting System which helps business firms to implement accounting process and makes it user friendly with automation.

3. Who is called Insolvent Person?

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Solution: Whose capital A/c shows debit balance and who is not in a position to meet his capital deficiency even from his private property is called an insolvent person.

4. What is Reserve Capital?

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Solution: Reserve Capital is that part of the subscribed capital which is reserved to be called-up only at the time of winding up or liquidation of the company.

5. What is Revaluation Account?

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Solution: The account which shows change in the values of assets and liabilities during the admission, retirement or death of a partner is known as Revaluation Account.

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(D) Complete the sentences: (5)

1. Partnership deed is an _______ of partnership.

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Solution: Partnership Deed is an Article of Partnership.

2. Aurangabad University prepares _______ Account instead of Profit and Loss Account.

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Solution: Aurangabad University prepares Income and Expenditure Account instead of Profit and Loss Account.

3. Returns outward are deducted from _______.

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Solution: Returns outward are deducted from Purchases.

4. New Ratio (–) _______ = Gain Ratio.

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Solution: New Ratio (less) Old ratio = Gain ratio

5. Cash receipts which are recurring in nature are called as _______ receipts.

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Solution: Cash receipts which are recurring in nature are called as revenue Receipts.

Q.2. Ram and Shyam were in partnership sharing profits and losses in the proportion of \(3:1\) respectively. Their Balance sheet as on 31st March, 2020 stood as follows:

Balance Sheet as on 31st March, 2020

They admit Bharat into partnership on 1st April, 2020. The term being that:

1. He shall have to bring in ₹ 40,000 as his Capital for \(\frac{1}{5}\)th share in future profit and ₹ 20,000 as his share of Goodwill.
2. A provision for 5% doubtful debts to be created on sundry debtors.
3. Stock should be appreciated by 5% and Land and Building be appreciated by 20%.
4. Furniture to be depreciated by 20%.
5. Capital Accounts of all partners be adjusted in their new profit sharing ratio through Cash Account.

Prepare:

• (i) Profit and Loss Adjustment Account
• (ii) Partners’ Capital Account.
• (iii) Balance Sheet of the new firm.

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SOLUTION:

OR

Ajay, Vijay and Sanjay were partners sharing profits and losses in the ratio of \(3 : 3 : 2\). Their Balance Sheet as on 31st March, 2020 is as follows:

Balance Sheet as on 31st March, 2020

On 1st April, 2020 Sanjay retired from the firm on the following terms:

1. R.D.D. is to be maintained at 10% on debtors.
2. ₹ 300 to be written off from creditors.
3. Goodwill of the firm is to be valued at ₹ 12,000, however, only Sanjay’s share in it is to be raised in the book and written off immediately.
4. Assets to be revalued as: Stock ₹ 18,900, Plant and Machinery ₹ 60,000, Live Stock ₹ 30,600.
5. The amount payable to Sanjay to be transferred to his loan account after retirement.

Prepare:

• (i) Revaluation Account.
• (ii) Partners’ Capital Account
• (iii) Balance Sheet of the New firm.

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SOLUTION:

Q.3. Asha, Usha and Nisha are partners in the firm sharing profits and losses in the ratio of \(3 : 2 : 1\) respectively. On 31st March, 2019 they decided to dissolve the firm when their Balance Sheet was as under:

Balance Sheet as on 31st March, 2019

The firm was dissolved on the above date and the assets realised as under:

1. Asha agreed to take over the Building at ₹ 1,23,600.
2. Usha took over Goodwill, Stock and Debtors at book value and agreed to pay Creditors and Bills payable.
3. Motor car and Machinery realised at ₹ 1,51,080 and ₹ 31,680 respectively.
4. Investment were taken by Nisha at an agreed value of ₹ 55,440.
5. Realisation Expenses amounted to ₹ 6,800.

Prepare:

• (a) Realisation Account
• (b) Partners’ Capital Account
• (c) Bank Account

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SOLUTION:

OR

Sonali draws a bill on Rupali for ₹ 50,000 for 3 months. Rupali accepts the bill on the same date. Sonali sends the bill to the bank for collection. Before due date, Rupali finds herself unable to make payment of bill and requests Sonali to renew it. Sonali agrees to the proposal on a condition that Rupali should pay ₹ 20,000 in cash along with interest ₹ 1,000 and accept a new bill for 2 months for the balance. Rupali retired the bill by paying ₹ 27,000.

Give Journal entries in the books of Sonali and prepare Rupali’s Account in the books of Sonali.

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SOLUTION:

Q.4. Ajita Ltd. issued 2,00,000 equity shares of ₹ 10 each at a premium of ₹ 2 per share payable as:

• ₹ 3 on application
• ₹ 5 on allotment (including ₹ 2 premium)
• ₹ 4 on first and final call

Applications were received for 2,40,000 equity shares and pro-rata allotment was made to all the applicants.

The excess application money was adjusted with allotment. Prerna who was allotted 400 shares failed to pay first and final call and her shares were forfeited.

Pass Journal Entries in the books of Ajita Ltd.

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SOLUTION:

OR

State the difference between Manual Accounting Process and Computerised Accounting Process.

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SOLUTION:

Basis of Difference	Manual Accounting	Computerized Accounting
1. Meaning	Manual accounting is the system in which we maintain physical register of journal and ledger for keeping the records of each business transaction.	In this system of computerized accounting, we use computer and different accounting software for digital record of each business transactions.
2. Calculation make total of	In this system, all calculations are done manually. For example, to find the balance of any ledger account. We will make total of the debit and credit side and then we will find its difference for showing balance.	In computerized accounting system, our duty is to record the business transactions manually in the database. All the calculations are done by computer system. We need not calculate each account’s balance, it is calculated automatically by computerized accounting system.
3. Ledger Accounts	Ledger accounts are prepared by posting transactions in appropriate ledger manually with the help of journal. There may be mistakes while transferring the amount manually.	In computerized accounting system, once a voucher is entered it will automatically be printed. Thus there is no chance of taking or transferring wrong amount.
4. Trial Balance	In this system of accounting, we have to take the balances of all ledger A/c, in Trial Balance Statement.	Computerized accounting system will produce Trial Balance automatically.
5. Adjustment Entries Record	Both adjustment journal entries and its posting in the ledger accounts will be done manually one by one.	Only adjustment entries will be passed in the computerized accounting system, posting in the Ledger accounts will be done automatically.
6. Financial Statements	We have to make the financial statements manually by carefully transferring Trial Balance’s figures into Trading, Profit and Loss Account and Balance Sheet.	We need not prepare financial statement manually; financial statements will be generated automatically. It will also automatically change after each voucher entry in the system. This facility is not available in the manual accounting system.
7. Closing the Books	After the year end accountants prepare financial statements for the accounting period. The balances are to be carried forward manually, to next year.	In computerized accounting software financial reports are auto-generated for the accounting period. The balances are automatically carried forward to next year.

Q.5. Anil, Sunil and Mohit were partners sharing profits and losses in the proportion of their capital. Their Balance Sheet as on 31st March, 2019 was as follows:

Balance Sheet as on 31st March, 2019

Mohit died on 1st August, 2019 and the following adjustments were made:

1. Assets to be revalued as under:
   • Land and Building ₹ 88,000
   • Motor Lorry ₹ 36,000
   • Furniture ₹ 34,000
2. All debtors were good.
3. Goodwill of the firm valued at two times the average profit of last 4 years’ profit.
4. Mohit’s share of profit to be calculated on the basis of average profit of the last three years.
5. Profit for four years 1st year ₹ 12,000, 2nd year ₹ 24,000, 3rd year ₹ 14,000, 4th year ₹ 22,000.

Prepare:

• (a) Mohit’s capital account showing the amount payable to his executor.
• (b) Give working note of Mohit’s share of goodwill and profit up to the date of his death

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SOLUTION:

OR

Following is the Balance Sheet of Param Company Ltd. as on 31st March, 2019 and 31st March, 2020:

You are required to prepare Comparative Balance Sheet of Param Company Ltd. as on 31st March, 2019 and 31st March 2020.

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SOLUTION:

Q.6. From the following Receipts and Payments Account of Shahu College, Kolhapur for the year ending 31st March, 2020 and additional information, prepare Income and Expenditure Account for the year ended 31st March, 2020 and Balance Sheet as on that date:

Receipts and Payments Account for the year ended 31st March, 2020

Additional information:

1. Outstanding Salaries ₹ 35,000.
2. 60% of donations are for Building Fund and Balance is to be treated as revenue income.

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SOLUTION:

Q.7. Asha and Nisha are partners sharing profits and losses in equal ratio. From the following Trial Balance and adjustments you are required to prepare Final Accounts:

Trial Balance as on 31st March, 2019

Adjustments:

1. Closing stock is valued at cost price ₹ 88,000 and market price ₹ 90,000.
2. Asha and Nisha withdrew goods from business ₹ 3,000 and ₹ 2,000 respectively for their personal use.
3. Depreciate Motor Van by 5% and Plant and Machinery by 7%.
4. Reserve for Doubtful debts on Debtors at 5% is to be created.
5. Outstanding Wages ₹ 800.

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SOLUTION:

Original Question Paper

Difficult Words and Their Meanings:

• Net Profit/Loss: The final financial gain or loss of a business after all operating expenses, interest, and taxes are subtracted from total revenues.
• Trading Account: A financial statement prepared to determine the gross profit or gross loss of a business from its primary trading activities (buying and selling goods).
• Profit and Loss Account: (Also Income Statement) A financial statement that summarizes revenues, costs, and expenses incurred during a specific period, to show the net profit or net loss.
• Financial Statement Analysis: The process of reviewing a company's financial statements (like Balance Sheet, Profit & Loss Account) to understand its financial health and make informed decisions.
• Creditors: Individuals or businesses to whom a company owes money for goods or services received on credit.
• Liquidity: A company's ability to meet its short-term financial obligations (debts) as they come due.
• Dissolution (of a firm): The formal process of ending a partnership or company, involving selling assets and settling liabilities.
• Admission (of a partner): The act of a new partner joining an existing partnership firm.
• Retirement (of a partner): The act of an existing partner leaving the partnership firm.
• Winding up: The process of closing down a company, selling its assets, paying off its debts, and distributing any remaining assets to the owners.
• Due Date (of a bill): The specific date on which payment for a bill of exchange or promissory note must be made.
• Debited: An entry on the left side of an account, typically increasing assets or expenses, or decreasing liabilities or equity.
• Credited: An entry on the right side of an account, typically increasing liabilities, equity or income, or decreasing assets or expenses.
• Profit and Loss Adjustment Account / Revaluation Account: An account prepared to record changes in the values of assets and liabilities when a partnership is reconstituted (e.g., admission, retirement, death of a partner).
• Gross Loss: Occurs when the cost of goods sold is higher than the net sales revenue in the Trading Account.
• Realization Expenses: Costs incurred during the process of dissolving a firm, such as expenses for selling assets and settling liabilities.
• Sacrifice Ratio: The ratio in which existing partners agree to give up (sacrifice) their share of profits in favor of a new incoming partner.
• Notary Public: A public official authorized to witness signatures, certify documents, and perform other legal formalities, such as noting the dishonor of a bill.
• Specific Donation: A donation received by an organization that must be used for a particular purpose specified by the donor.
• Capital Receipt: Money received by a business that is not from its regular operations (e.g., sale of a fixed asset, long-term loan). It's not treated as income for the current period.
• Legacy: A gift of money or property left to an individual or organization through a will after the donor's death.
• Computerized Accounting System (CAS): Using computer hardware and software to record, process, and report financial transactions.
• Insolvent Person: An individual or entity whose liabilities (debts) exceed their assets, making them unable to pay their debts.
• Reserve Capital: A portion of a company's uncalled share capital that is set aside to be called up only in the event of the company's winding up.
• Partnership Deed: A written agreement among partners outlining the terms, conditions, rights, and responsibilities of the partnership.
• Income and Expenditure Account: A summary of all income and expenses for a specific period, prepared by non-profit organizations to determine surplus or deficit.
• Returns Outward: (Purchase Returns) Goods returned by a business to its suppliers.
• Gain Ratio: The ratio in which the continuing partners benefit from the share of profit of an outgoing (retiring or deceased) partner.
• Revenue Receipts: Money received by a business from its normal day-to-day operations (e.g., sales, service fees).
• Goodwill: An intangible asset representing the non-physical value of a business, such as its reputation, customer loyalty, and brand recognition.
• Provision for Doubtful Debts (R.D.D.): An estimated amount set aside from profits to cover potential losses from customers who may not pay their debts.
• Sundry Debtors: Customers who owe money to the business for goods or services supplied on credit.
• Appreciated (assets): An increase in the value of an asset over time.
• Depreciated (assets): A decrease in the value of an asset over time due to use, wear and tear, or obsolescence.
• Pro-rata Allotment: Allocation of shares to applicants in proportion to their application when a share issue is oversubscribed.
• Forfeited Shares: Shares that are taken back by the company from a shareholder who fails to pay allotment or call money due on them.